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Date May 2014 Marks available 2 Reference code 14M.2.sl.TZ2.6
Level SL only Paper 2 Time zone TZ2
Command term Calculate Question number 6 Adapted from N/A

Question

The front view of the edge of a water tank is drawn on a set of axes shown below.

The edge is modelled by \(y = a{x^2} + c\).



Point \({\text{P}}\) has coordinates \((-3, 1.8)\), point \({\text{O}}\) has coordinates \((0, 0)\) and point \({\text{Q}}\) has coordinates \((3, 1.8)\).

Write down the value of \(c\).

[1]
a.

Find the value of \(a\).

[2]
b.

Hence write down the equation of the quadratic function which models the edge of the water tank.

[1]
c.

The water tank is shown below. It is partially filled with water.


Calculate the value of y when \(x = 2.4{\text{ m}}\).

[2]
d.

The water tank is shown below. It is partially filled with water.


State what the value of \(x\) and the value of \(y\) represent for this water tank.

[2]
e.

The water tank is shown below. It is partially filled with water.


Find the value of \(x\) when the height of water in the tank is \(0.9\) m.

[2]
f.

The water tank is shown below. It is partially filled with water.


 

The water tank has a length of 5 m.

 

When the water tank is filled to a height of \(0.9\) m, the front cross-sectional area of the water is \({\text{2.55 }}{{\text{m}}^2}\).

(i)     Calculate the volume of water in the tank.

The total volume of the tank is \({\text{36 }}{{\text{m}}^3}\).

(ii)     Calculate the percentage of water in the tank.

[2]
g.

Markscheme

\(0\)     (A1)(G1)

[1 mark]

a.

\(1.8 = a{(3)^2} + 0\)     (M1)

OR

\(1.8 = a{( - 3)^2} + 0\)     (M1)

 

Note: Award (M1) for substitution of \(y = 1.8\) or \(x = 3\) and their value of \(c\) into equation. \(0\) may be implied.

 

\(a = 0.2\)   \(\left( {\frac{1}{5}} \right)\)     (A1)(ft)(G1)

 

Note: Follow through from their answer to part (a).

     Award (G1) for a correct answer only.

 

[2 marks]

b.

\(y = 0.2{x^2}\)     (A1)(ft)

 

Note: Follow through from their answers to parts (a) and (b).

     Answer must be an equation.

 

[1 mark]

c.

\(0.2 \times {(2.4)^2}\)     (M1)

\( = 1.15{\text{ (m)}}\)   \((1.152)\)     (A1)(ft)(G1)

 

Notes: Award (M1) for correctly substituted formula, (A1) for correct answer. Follow through from their answer to part (c).

     Award (G1) for a correct answer only.

 

[2 marks]

d.

\(y\) is the height     (A1)

positive value of \(x\) is half the width (or equivalent)     (A1)

[2 marks]

e.

\(0.9 = 0.2{x^2}\)     (M1)

 

Note: Award (M1) for setting their equation equal to \(0.9\).

 

\(x =  \pm 2.12{\text{ (m)}}\)   \(\left( { \pm \frac{3}{2}\sqrt 2 ,{\text{ }} \pm \sqrt {4.5} ,{\text{ }} \pm {\text{2.12132}} \ldots } \right)\)     (A1)(ft)(G1)

 

Note: Accept \(2.12\). Award (G1) for a correct answer only.

 

[2 marks]

f.

(i)     \(2.55 \times 5\)     (M1)

 

Note: Award (M1) for correct substitution in formula.

 

\( = 12.8{\text{ (}}{{\text{m}}^3}{\text{)}}\)   \(\left( {{\text{12.75 (}}{{\text{m}}^3}{\text{)}}} \right)\)     (A1)(G2)

[2 marks]

 

(ii)     \(\frac{{12.75}}{{36}} \times 100\)     (M1)

 

Note: Award (M1) for correct quotient multiplied by \(100\).

 

\( = 35.4 (\%)\)  \((35.4166 \ldots )\)     (A1)(ft)(G2)

 

Note: Award (G2) for \(35.6 (\%) (35.5555… (\%))\).

     Follow through from their answer to part (g)(i).

 

[2 marks]

g.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.
[N/A]
g.

Syllabus sections

Topic 6 - Mathematical models » 6.3 » Quadratic models.
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